The paramagnetic behaviour of $B_2$ is due to the presence of

  • A
    two unpaired electrons in $\pi^*$ $MO$
  • B
    two unpaired electrons in $\pi$ $MO$
  • C
    two unpaired electrons in $\sigma$ $MO$
  • D
    two unpaired electrons in $\sigma^*$ $MO$

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Similar Questions

Match the following molecules/ions in List-$I$ with their respective bond orders in List-$II$:
| List-$I$ (Molecules/ions) | List-$II$ (Bond order) |
| :--- | :--- |
| $A. N_2^+$ | $I. 1.0$ |
| $B. CO$ | $II. 1.5$ |
| $C. O_2$ | $III. 2.0$ |
| $D. O_2^-$ | $IV. 2.5$ |
| | $V. 3.0$ |

The difference between bond orders of $CO$ and $NO^{\oplus}$ is $\frac{x}{2}$ where $x = .....$
(Round off to the Nearest Integer)

The pair of species that has the same bond order in the following is

Give the order of relative stability for $N_2, N_2^+, N_2^-,$ and $N_2^{2+}$.

After understanding the assertion and reason, choose the correct option.
Assertion : In the bonding molecular orbital $(MO)$ of $H_2,$ electron density is increased between the nuclei.
Reason : The bonding $MO$ is $\psi_A + \psi_B,$ which shows destructive interference of the combining electron waves.

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